Let a $11$x$11$- square , that have $121$ mini-squares, and the numbers from $1$ to $121$.
Show that they can't be arranged such that two consecutive numbers stay in two mini squares that have a common edge and the numbers $1^2$, $2^2$,...,$11^2$ stay on the same column.
I think that on the line that contains $10^2$ are numbers between $90$ and $110$.
Also on the line that contains $9^2$ are numbers between $71$ and $91$ and so on.
I don't know how it can help me.
On
When we start filling up the $7^{th}$ row (with parts of row $8$) we block off the green tiles. So we can not fill them up later (two consecutive numbers have to stay in two mini squares that have a common edge). Thus we can easily see that it can not be so that $2$ consecutive numbers stay in two mini squares that have a common edge and the numbers $1^2, 2^2,...,11^2$ stay on the same column. Hope you liked it!
P.S. I had no suitable tool for doing this so I had to do this in Microsoft Word. And to color the boxes I had to type $X$s and then highlight it (my Microsoft Excel needs to be updated). Hope this is not inconvenient for you.
Edit: Excel is back on. However understand that my presented case was the best case. Here I have included on exactly why and another which shows where you are bound to get stuck.
So the numbers $1^2$ through to $11^2$ fill one column. This divides the square into two parts. There are ten differences between consecutive squares, and these differences are all odd. There are an even number of distinct integers strictly between two consecutive squares.
Now at each square other than $1$ and $121$ the chain of successive digits must cross the line of squares eg $3\to 4\to 5$ and $8\to 9 \to 10$ (can't go vertically in the line of squares). Say we go left from $1$, then on the left we accumulate the differences $2, 6, 10, 14, 18$ which total $50$ and this is not divisible by $11$ - and the differences on the right are $4+8+12+16+20=60$ and this isn't divisible by $11$ either.